In electrochemistry, Gibbs free energy from EMF numericals mcqs one of the most important concepts to grasp is the relationship between the Gibbs free energy (ΔG\Delta GΔG) and the electrochemical cell potential (EEE). This connection is fundamental for understanding how energy is transferred and stored in electrochemical systems. The relationship between Gibbs free energy and EMF (electromotive force) can be understood through the equation:

ΔG=−nFE\Delta G = -nFEΔG=−nFE

where:

• ΔG\Delta GΔG is the Gibbs free energy change,

• nnn is the number of moles of electrons transferred in the reaction,

• FFF is the Faraday constant (approximately 96500 C/mol),

• EEE is the cell potential.

The Role of Gibbs Free Energy in Electrochemical Reactions

In Gibbs free energy from EMF numericals mcqs, the concept of Gibbs free energy is critical because it helps determine whether a chemical reaction is spontaneous. For an electrochemical reaction to be spontaneous, the Gibbs free energy must be negative. This implies that a positive EMF is required for the reaction to occur spontaneously. Conversely, if the Gibbs free energy is positive, the reaction is non-spontaneous, and a negative EMF indicates that the reaction cannot proceed without external energy.

How to Calculate Gibbs Free Energy from EMF

To calculate the Gibbs free energy from the EMF, the cell potential must first be determined. The equation ΔG=−nFE\Delta G = -nFEΔG=−nFE directly links the free energy change to the cell potential. This relationship allows us to calculate how much useful work can be derived from the electrochemical reaction. For example, if we know the cell potential and the number of moles of electrons transferred, we can easily calculate the Gibbs free energy and, consequently, predict whether the reaction can proceed spontaneously.

Example 1: A Simple Calculation of Gibbs Free Energy from EMF Numericals MCQs

Let’s take an example to calculate Gibbs free energy from the EMF of a galvanic cell. For a given cell:

Zn(s)+Cu2+(aq)→Zn2+(aq)+Cu(s)\text{Zn} (s) + \text{Cu}^{2+} (aq) \rightarrow \text{Zn}^{2+} (aq) + \text{Cu} (s)Zn(s)+Cu2+(aq)→Zn2+(aq)+Cu(s)

Assume the cell potential (EEE) is 1.10 V, the number of moles of electrons transferred is 2, and the temperature is 298 K. The Faraday constant is 96500 C/mol. The Gibbs free energy can be calculated using the formula:

ΔG=−nFE\Delta G = -nFEΔG=−nFE

Substituting the values:

ΔG=−2×96500×1.10=−212300 J/mol\Delta G = -2 \times 96500 \times 1.10 = -212300 \, \text{J/mol}ΔG=−2×96500×1.10=−212300J/mol

This negative value indicates that the reaction is spontaneous under standard conditions.

Understanding the Link Between Gibbs Free Energy and EMF

The relationship between Gibbs free energy and EMF not only helps in understanding spontaneity but also in calculating the efficiency of a battery or fuel cell. The amount of electrical energy generated by an electrochemical reaction can be directly related to the change in Gibbs free energy. Therefore, calculating Gibbs free energy from EMF numericals mcqs is essential for determining the practical application and the energy yield of a system.

Example 2: Non-Spontaneous Reactions

Let’s now consider a non-spontaneous reaction where the cell potential is negative. For such a reaction, the Gibbs free energy will also be positive. This means that to drive the reaction, we would need to apply an external electrical force (such as in electrolysis). For example, if the cell potential is -1.5 V, and the number of moles of electrons transferred is 3, the Gibbs free energy will be:

ΔG=−nFE=−3×96500×(−1.5)=434250 J/mol\Delta G = -nFE = -3 \times 96500 \times (-1.5) = 434250 \, \text{J/mol}ΔG=−nFE=−3×96500×(−1.5)=434250J/mol

Here, the positive Gibbs free energy confirms that the reaction is non-spontaneous and requires external work.

Practical Applications of Gibbs Free Energy and EMF

Understanding how to calculate Gibbs free energy from EMF numericals mcqs is pivotal in various applications. In battery technology, knowing the Gibbs free energy helps in evaluating the energy that can be harvested from a chemical reaction. It is also crucial in electrolysis, where the electrochemical cells are used to drive non-spontaneous reactions, such as the decomposition of water into hydrogen and oxygen.

Similarly, in fuel cells, understanding the Gibbs free energy from EMF numericals mcqs helps determine the potential efficiency and power output. Fuel cells rely on spontaneous redox reactions, and the higher the EMF, the greater the potential energy available for useful work.

Gibbs Free Energy from EMF Numericals MCQs

1.  For the reaction taking place in the cell:
   Pt(s) | H₂(g) | H⁺(aq) || Ag⁺(aq) | Ag(s)
   Ecell0E^0_{cell}Ecell0​ = +0.5332V. The value of ΔG° is _____ kJ mol⁻¹ [in nearest integer]
   a) 10
   b) 5
   c) 50
   d) 51
   Answer: d

2. The cell potential for the given cell at 298 K
   Pt|H₂(g) (1 bar) | H⁺(aq)(1M) || Cu²⁺(aq)|(Cu) is 0.31V. The pH of the acidic solution is found to be 3, whereas the concentration of Cu²⁺ is 10⁻ˣ M. The value of x is _____
   a) 7
   b) 3
   c) 6
   d) 9
   Answer: a

3. Cu(s) + Sn²⁺ (0.001 M) → Cu²⁺ (0.01M) + Sn(s). The Gibbs free energy change for the above reaction at 298 K is X * 10⁻¹ kJ mol⁻¹. The value of X is _____ [Nearest integer]
   a) 700
   b) 883
   c) 900
   d) 983
   Answer: d

4. In a cell, the following reactions take place:
   Fe²⁺ → Fe³⁺ + e⁻, EFe3+/Fe2+0E^0_{{Fe}³⁺/{Fe}²⁺}EFe3+/Fe2+0​ = 0.77V
   2I⁻ → I₂ + 2e⁻ EI2/I−0E^0_{{I_2}/{I}^-}EI2​/I−0​ = 0.54V
   The standard electrode potential for the spontaneous reaction in the cell is X * 10⁻² V at 298 K. The value of X is _____ [Nearest integer]
   a) 23
   b) 20
   c) 3
   d) 33
   Answer: a

5. A diluted solution of sulphuric acid is electrolyzed using a current of 0.10 A for 2 hours to produce hydrogen and oxygen gas. The total volume of gases produced at STP is _____ cm³ (Nearest integer)
   a) 1
   b) 12
   c) 127
   d) 120
   Answer: c

6. The limiting molar conductivities of NaI, NaNO₃, and AgNO₃ are 12.7, 12.0, and 13.3 mS m² mol⁻¹ respectively (all at 25°C). The limiting molar conductivity of AgI at this temperature is _____ mS m² mol⁻¹
   a) 14
   b) 1.4
   c) 10
   d) 1
   Answer: a

7. The cell potential for the following cell
   Pt |H₂(g)|H⁺(aq) || Cu²⁺(0.01M)|Cu(s)
   is 0.576 V at 298 K. The pH of the solution is _____ (Nearest integer)
   a) 1
   b) 2
   c) 3
   d) 5
   Answer: d

8. The quantity of electricity of Faraday needed to reduce 1 mole of Cr₂O₇⁻² to Cr³⁺ is
   a) 5
   b) 6
   c) 7
   d) 8
   Answer: b

9. For the given reactions
   Sn²⁺ + 2e⁻ → Sn
   Sn⁴⁺ + 4e⁻ → Sn
   The potential are for
   ESn2+/Sn0E^0_{{Sn}²⁺/{Sn}}ESn2+/Sn0​ = -0.140 V and ESn+/Sn0E^0_{{Sn}^4⁺/{Sn}}ESn+/Sn0​ = 0.010 V
   The magnitude of standard potential for Sn⁴⁺/Sn²⁺ i.e. ESn+/Sn2+0E^0_{{Sn}^4⁺/{Sn}^{2+}}ESn+/Sn2+0​ is ____ * 10⁻² V. (Nearest integer)
   a) 20
   b) 5
   c) 16
   d) 10
   Answer: c

10. The (∂E∂T)P\left(\frac{\partial E}{\partial T}\right)_P(∂T∂E​)P​ of different types of half cells are as follows:
    A B C D
    110⁻⁴ 210⁻⁴ 0.110⁻⁴ 0.210⁻⁴
    (Where E is the electromotive force).
    Which of the above half cells would be preferred to be used as reference electrode?
    a) A
    b) B
    c) C
    d) D
    Answer: c

11. The resistance of a conductivity cell containing 0.01 M KCl solution at 298 K is 1750 Ω. If the conductivity of 0.01 M KCl solution at 298 K is 0.152 * 10⁻³ S cm⁻¹, then the cell constant of the conductivity cell is _____ * 10⁻³ cm⁻¹.
    a) 300
    b) 366
    c) 200
    d) 266
    Answer: d

12. In Fuel cells _____ are used as catalysts.
    a) Zinc – Mercury
    b) Lead – Manganese
    c) Platinum – Palladium
    d) Nickel – Cadmium
    Answer: c

13. The molar conductivity is maximum for the solution of concentration
    a) 0.005M
    b) 0.001M
    c) 0.004M
    d) 0.002M
    Answer: b

14. Half-life of a reaction is found to be inversely proportional to the fifth power in initial concentration, the order of reaction is
    a) 5
    b) 6
    c) 3
    d) 4
    Answer: b

15. The correct order of reduction potentials of the following pairs is
    (A) Cl₂/Cl⁻
    (B) I₂/I⁻
    (C) Ag⁺/Ag
    (D) Na⁺/Na
    (E) Li⁺/Li
    Choose the correct answer from the options given below
    a) A > C > B > D > E
    b) E > C > B > D > A
    c) E > A > B > D > C
    d) E > A > D > B > C
    Answer: a

16. 96.5 amperes current is passed through the molten AlCl₃ for 100 seconds. The mass of aluminum deposited at the cathode is (Atomic weight of Al = 27 u)
    a) 0.90 g
    b) 0.45 g
    c) 1.35 g
    d) 1.80 g
    Answer: a

17. The E° of Ce⁴⁺/Ce³⁺ is 1.60 V and that of Fe³⁺/Fe²⁺ is 0.76 V. The E° for Fe³⁺ oxidizing Ce³⁺ is
    a) (+0.84 V)
    b) -0.84 V
    c) (+2.32 V)
    d) -2.32 V
    Answer: b

18. Given the E° for Mn⁷⁺/Mn²⁺ = 1.51V, E° for Mn⁴⁺/Mn²⁺ = 1.23V.
    Calculate the EMn+/Mn4+0E^0_{{Mn}^7⁺/{Mn}^{4+}}EMn+/Mn4+0​
    a) 1.7 V
    b) 0.48 V
    c) 0.28 V
    d) -0.28 V
    Answer: a

19. The cell potential, EcellE_{cell}Ecell​, for the following cell notation (in Volts) is
    A(s)|A⁺(aq, 0.1M) || B²⁺(aq, 0.01M)|B(s)
    EA+/A0E^0_{{A}⁺/{A}}EA+/A0​ = 1 V EB2+/B0E^0_{{B}²⁺/{B}}EB2+/B0​ = 3 V
    a) 4
    b) 3
    c) 2
    d) 1
    Answer: c

20. The standard reduction potentials at 298 K for the following half cell reactions are given below:
    Zn²⁺(aq) + 2e⁻ ⇌ Zn(s), -0.762 V
    Cu³⁺(aq) + 3e⁻ ⇌ Cr(s), -0.74 V
    2H⁺(aq) + 2e⁻ ⇌ H₂(g), 0.00 V
    Fe³⁺(aq) + e⁻ ⇌ Fe²⁺(aq), +0.77 V
    Which one of the following is the strongest reducing agent?
    a) Zn
    b) Cr
    c) H₂
    d) Fe²⁺
    Answer: a

21. The standard reduction potentials for Zn²⁺ / Zn, Ni²⁺ / Ni and Fe²⁺ / Fe are -0.76, -0.23 and -0.44 V, respectively. The reaction X + Y²⁺ → X²⁺ + Y will be spontaneous when
    a) X = Ni, Y = Fe
    b) X = Ni, Y = Zn
    c) X = Fe, Y = Zn
    d) X = Zn, Y = Ni
    Answer: d

22. During the electrolysis of copper sulphate aqueous solution using copper electrode, the reaction taking place at the cathode is
    a) Cu → Cu²⁺ (aq) + 2e⁻
    b) Cu²⁺ (aq) + 2e⁻ → Cu(s)
    c) H⁺ (aq) + e⁻ → 12\frac{1}{2}21​ H₂ (g)
    d) SO₄⁻² (aq) → SO₃ (g) + 12\frac{1}{2}21​ O₂ (g) + 2e⁻
    Answer: b

23. For the following cell reaction
    Ag|Ag⁺ / AgCl | Cl⁻|Cl₂, Pt
    ΔGf0(AgCl)\Delta G^0_f(AgCl)ΔGf0​(AgCl) = -109 kJ/mol
    ΔGf0(Cl−)\Delta G^0_f(Cl^-)ΔGf0​(Cl−) = -129 kJ/mol
    ΔGf0(Ag+)\Delta G^0_f(Ag^+)ΔGf0​(Ag+) = 78 kJ/mol
    E⁰ of the cell is
    a) -0.6 V
    b) 0.6 V
    c) 6 V
    d) None of these
    Answer: a

24. Corrosion of iron is essentially an electrochemical phenomenon where the cell reactions are
    a) Fe is oxidized to Fe²⁺ and dissolved oxygen in water is reduced to OH⁻
    b) Fe is oxidized to Fe³⁺ and H₂O is reduced to O₂⁻
    c) Fe is oxidized to Fe²⁺ and H₂O is reduced to O₂⁻
    d) Fe is oxidized to Fe²⁺ and H₂O is reduced to O₂
    Answer: a

25. For cell reaction Zn + Cu²⁺ → Zn²⁺ + Cu , cell representation is
    a) Zn|Zn²⁺||Cu²⁺|Cu
    b) Cu|Cu²⁺||Zn²⁺|Zn
    c) Cu|Zn²⁺||Zn|Cu²⁺
    d) Cu²⁺|Zn||Zn²⁺|Cu
    Answer: a

26. For the redox reaction
    Zn(s) + Cu²⁺ (0.1M) → Zn²⁺ (1M) + Cu(s)
    taking place in a cell Ecell0 E^0_{cell}Ecell0​ is 1.10 V. Ecell E_{cell}Ecell​ for the cell will be
    ( 2.303RTF\frac{2.303RT}{F}F2.303RT​ = 0.0591)
    a) 2.14 V
    b) 1.80 V
    c) 0.82 V
    d) 1.07 V
    Answer: d

27. The standard electrode potential of three metals X, Y, and Z are -1.2 V, +0.5 V, and -3.0 V respectively. The reducing power of these metals will be
    a) Y > X > Z
    b) Y > Z > X
    c) Y > Z > X
    d) Z > X > Y
    Answer: d

28. Consider the following electrodes:
    P = Zn²⁺ (0.0001 M)/Zn
    Q = Zn²⁺ (0.1 M)/Zn
    R = Zn²⁺ (0.01 M)/Zn
    S = Zn²⁺ (0.001 M)/Zn
    EZn/Zn2+0E^0_{{Zn}/{Zn}²⁺}EZn/Zn2+0​ = -0.76 V
    Electrode potentials of the above electrodes in volts are in the order
    a) P > R > S > Q
    b) Q > P > S > R
    c) R > S > P > Q
    d) Q > R > S > P
    Answer: d

29. In the electrochemical cell:
    Zn/ZnSO₄(0.01M)||CuSO₄(1.0M)|Cu , the emf of this Daniell cell is E₁ . When the concentration of ZnSO₄ is changed to 1.0 M and that of CuSO₄ changed to 0.01 M, the emf changes to E₂. From the following which one is the relationship between E₁ and E₂ ? (Given, RT/F = 0.059)
    a) None of these
    b) E₁ > E₂
    c) E₂ > E₁
    d) E₂ = E₁
    Answer: b

Conclusion on Gibbs Free Energy from EMF Numericals MCQs

In conclusion, the ability to calculate Gibbs free energy from EMF numericals mcqs is a foundational concept in electrochemistry that helps aspirants understand the feasibility and efficiency of electrochemical reactions. By applying the formula ΔG=−nFE\Delta G = -nFEΔG=−nFE, aspirants can predict whether a reaction will be spontaneous and calculate the amount of work that can be extracted from the reaction. The concept of Gibbs free energy is central to the operation of electrochemical cells, and it plays a critical role in the development of energy storage systems, fuel cells, and various other electrochemical applications. The link between Gibbs free energy and EMF is not just theoretical but has direct practical implications in modern technology.

Solving Gibbs free energy from EMF numericals mcqs strengthens conceptual claritty for class 12 chemistry and competitive exams like NEET, JEE and CUET. Therefore, practicing Gibbs free energy from EMF numericals mcqs is essential for mastering numerical problems based on electrochemical cells.